Abstracted StringDivision to work with arbitrary containers and digit-types
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@@ -1,52 +1,5 @@
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#include "GeneralUtility.h"
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#include <string>
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#include <sstream>
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#include <stdexcept>
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// Based on: https://www.geeksforgeeks.org/divide-large-number-represented-string/
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std::pair<std::string, int>
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GeneralUtility::StringDivision(const std::string& dividend, const unsigned int divisor, const std::string& set) {
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// No set? Throw logic error
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if (set.length() == 0)
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throw std::logic_error("Can't divide a number of base0! Please supply a nonempty set!");
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// No division by 0
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if (divisor == 0)
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throw std::overflow_error("Division by zero!");
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// As result can be very large store it in string
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std::stringstream ss;
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// Find prefix of number that is larger than divisor.
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int idx = 0;
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int temp = Ord(dividend[idx], set);
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while (temp < divisor)
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temp = temp * set.length() + Ord(dividend[++idx], set);
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// Repeatedly divide divisor with temp. After
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// every division, update temp to include one
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// more digit.
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int curRest = temp % divisor;
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while (dividend.size() > idx) {
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// Store result in answer i.e. temp / divisor
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ss << (char)(set[(temp / divisor)]);
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curRest = temp % divisor;
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// Take next digit of number
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temp = (temp % divisor) * set.length() + Ord(dividend[++idx], set);
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}
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// If divisor is greater than number
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if (ss.str().length() == 0)
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{
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// Generate 0-value string
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std::stringstream ss;
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ss << set[0];
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return std::make_pair(ss.str(), BaseX_2_10(dividend, set));
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}
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// else return the answer
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return std::make_pair(ss.str(), curRest);
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}
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std::string GeneralUtility::BaseX_2_Y(const std::string &num, const std::string &set_in, const std::string &set_out, const std::uint32_t minOutLen) {
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if ((set_in.length() == 0) || (set_out.length() == 0))
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@@ -64,7 +17,7 @@ std::string GeneralUtility::BaseX_2_Y(const std::string &num, const std::string
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std::string buf = num;
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while (buf != zeroInbase) {
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const auto divRes = StringDivision(buf, set_out.length(), set_in);
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const auto divRes = DigitstringDivision(buf, set_out.length(), set_in);
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const std::uint64_t mod = divRes.second;
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buf = divRes.first;
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ss << set_out[mod];
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@@ -3,6 +3,8 @@
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#include <algorithm>
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#include <utility>
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#include <sstream>
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#include <stdexcept>
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class GeneralUtility {
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public:
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@@ -21,7 +23,8 @@ public:
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//! \param divisor The integer divisor
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//! \param set The set/base of `dividend`
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//! \return A pair of the result. (result, rest)
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static std::pair<std::string, int> StringDivision(const std::string& dividend, const unsigned int divisor, const std::string& set = "0123456789");
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template <class T_Container>
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static std::pair<T_Container, int> DigitstringDivision(const T_Container& dividend, const unsigned int divisor, const T_Container& set);
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//! Will convert a number of arbitrary base to base 10
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//! \param num A string representing the number
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@@ -95,4 +98,69 @@ int GeneralUtility::Ord(const T_Type& item, const T_Container& set) {
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return result - set.begin();
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}
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// Based on: https://www.geeksforgeeks.org/divide-large-number-represented-string/
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template <class T_Container>
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std::pair<T_Container, int>
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GeneralUtility::DigitstringDivision(const T_Container& dividend, const unsigned int divisor, const T_Container& set) {
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// Quick rejects:
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// No set? Throw logic error
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if (set.size() == 0)
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throw std::logic_error("Can't divide a number of base0! Please supply a nonempty set!");
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// No division by 0
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if (divisor == 0)
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throw std::overflow_error("Division by zero!");
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// Dividend size 0? Return 0.
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if (dividend.size() == 0)
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return std::make_pair(T_Container({set[0]}), 0);
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// Verify that all digits are represented in the set/base
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for (const auto& c : dividend)
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if (Ord(c, set) == -1)
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throw std::logic_error("The supplied dividend contains a digit that is not represented in the set/base!");
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// Now for the actual algorithm:
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T_Container result;
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// Find prefix of number that is larger than divisor.
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int idx = 0;
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int temp = Ord(dividend[idx], set);
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while (temp < divisor) {
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idx++;
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if (idx < dividend.size())
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temp = temp * set.size() + Ord(dividend[idx], set);
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else
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break;
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}
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// Repeatedly divide divisor with temp. After
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// every division, update temp to include one
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// more digit.
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int curRest = temp % divisor;
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while (dividend.size() > idx) {
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// Store result in answer i.e. temp / divisor
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result.insert(result.cend(), set[temp / divisor]);
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curRest = temp % divisor;
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// Take next digit of number
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idx++;
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if (idx < dividend.size())
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temp = (temp % divisor) * set.size() + Ord(dividend[idx], set);
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}
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// If divisor is greater than number
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if (result.size() == 0) {
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// Generate 0-value digitstring
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result.clear();
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result.insert(result.cend(), set[0]);
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return std::make_pair(result, BaseX_2_10(dividend, set));
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}
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// else return the answer
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return std::make_pair(result, curRest);
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}
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#endif //GENERALUTILITY_GENERALUTILITY_H
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